The inside of the square root gives negative answers for part of your ##\theta## range. What part?
So split your outer integral into two. One for the part where the inside of the square root is positive, and another where it isn't.
What are the limits of ##\theta## for that split?

What should your lower bound for the inner integral be in your second inner integral (Hint: look at the diagram)?

The inside of the square root gives negative answers for part of your ##\theta## range. What part?
So split your outer integral into two. One for the part where the inside of the square root is positive, and another where it isn't.
What are the limits of ##\theta## for that split?

What should your lower bound for the inner integral be in your second inner integral (Hint: look at the diagram)?

I don't understand, where, and why is my square root giving negative values? And how am I splitting this up? I don't see where I went wrong.

Apparently, this is an incorrect answer... I think I am having more of a conceptual problem with the regions than anything else.

Look again at the diagram. For some values of ##\theta \in (0, \pi/2)## the ##r##-integration runs from some ##r_1(\theta)## to ##r_2(\theta)##, but for other values of ##\theta## it runs from ##r=0## to ##r = r_2(\theta)##.

The inside of the square root gives negative answers for part of your ##\theta## range. What part?
So split your outer integral into two. One for the part where the inside of the square root is positive, and another where it isn't.
What are the limits of ##\theta## for that split?

What should your lower bound for the inner integral be in your second inner integral (Hint: look at the diagram)?

Look again at the diagram. For some values of ##\theta \in (0, \pi/2)## the ##r##-integration runs from some ##r_1(\theta)## to ##r_2(\theta)##, but for other values of ##\theta## it runs from ##r=0## to ##r = r_2(\theta)##.

I apologize for the late response, but I'm sorry I still don't understand what you mean by when it goes negative? I've tried conceptualizing it.

The solution for this problem was also posted by my teacher, but I am still trying to grasp it.

In the first integral, I can understand where the ##\frac {\pi} {4} ## is coming from, as it's the upper bound of the leminscate, but I plotted the image, and the ##\frac {\pi} {6}## seems to correspond to nothing on the region, it seems arbitrary.

I also don't understand why the second integral changes its bounds to (pi/6, pi/2) all of a sudden... I'm just not understanding these bounds at all.

Apparently, this is an incorrect answer... I think I am having more of a conceptual problem with the regions than anything else.

Well, you are interested in only one part of the lemniscate which is located inside the white circle.

If you were to plot the equation for the lemniscate, starting at θ = 0, you would have a point located at approximately (14.14,0) on the positive x-axis. However, this point is located outside the region enclosed by the white circle.

What you want to do is calculate the angle at which the lemniscate and the circle intersect after θ = 0 and before θ = π/4; in other words, solve ##100 sin (θ) = \sqrt{200 ⋅ cos (2θ)}## for θ. This value of θ becomes the lower bound of your integration.

Well, you are interested in only one part of the lemniscate which is located inside the white circle.

If you were to plot the equation for the lemniscate, starting at θ = 0, you would have a point located at approximately (14.14,0) on the positive x-axis. However, this point is located outside the region enclosed by the white circle.

What you want to do is calculate the angle at which the lemniscate and the circle intersect after θ = 0 and before θ = π/4; in other words, solve ##100 sin (θ) = \sqrt{200 ⋅ cos (2θ)}## for θ. This value of θ becomes the lower bound of your integration.

Okay, thank you, it completely slipped my mind that the theta still traversed the leminscate even after pi/4. My only issue now is solving this equation, which although it is algebra, I'm not have a very good time with it.

Okay, thank you, it completely slipped my mind that the theta still traversed the leminscate even after pi/4. My only issue now is solving this equation, which although it is algebra, I'm not have a very good time with it.

To make things a little easier, there's a trig identity which you can use on cos (2θ) to convert it to an expression in sin (θ).